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A block of mass 2 kg is dropped from a h...

A block of mass 2 kg is dropped from a height of 0-4 mon a spring whose force constant is 1960 N/m. What will be the maximum distance x of the compression of the spring ?

A

0.5 m

B

0.1 m

C

0.01 m

D

1 m

Text Solution

Verified by Experts

The correct Answer is:
b
P.E. of spring = loss in P.E. of body
`1/2 kx^(2)=mg(h+x)`
`2(0.4+x)9.8=(1)/(2)1960xxx^(2)`
`19.6(0.4+x)=980 x^(2)`
Which gives on solving, the quadratic `x=(1)/(10)=0.1 m`
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Knowledge Check

  • A block of mass 'm' initially at rest is dropped from a height 'h' into a spring whose constant is k. The maximum distance through which the spring is compressed is :

    A
    `(2mgh)/(k)`
    B
    `(2mg(h+k))/(3)`
    C
    `(mg-sqrt(m^(2)g^(2)+2mghk))/(2k)`
    D
    `(mg+sqrt(m^(2)g^(2)+2mghk))/(k)`

  • A slab of mass 'm' is released from a height x to the top a spring of force constant k. The maximum compression of the spring is y. Then

    A
    `mg x = 1/2 ky^2`
    B
    `mg (x + y) = 1/2 ky^2`
    C
    `mg (x + y) = 1/2 kx^2`
    D
    `mg x = 1/2 x (x + y)^2`

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    B
    `T=2pi sqrt(((M+m)x)/(mg))`
    C
    `T=2pi sqrt(((M+m))/(mgx))`
    D
    `T=2pi sqrt((mgx)/((M+m)))`

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